Adding, Subtracting, Multiplying and Dividing Significant Figures

I observed in my classes that my students are always asking if how many decimal places are they going to round off the result of their calculations.  As a student they must know how to handle significant figures in doing calculations in addition, subtraction, multiplication and division.  You may have correct answer but the way to round off your answer to proper number of significant figures also matters.

There are rules to follow in using significant figures in calculations.
1.  In addition and subtraction, the calculated result must be rounded off to proper number of digits following the number which has least number of digits after the decimal point.   Let us have an example

                                                 90.335   cm   (3 digits after the decimal point)

                                           +      2.1       cm   (1 digit after the decimal point)
                                           ______________
                                                92.435   cm
                                     or        92.4  cm is the final answer.

The result is rounded off to one digit after the decimal point following the rule that in rounding off, answer in addition and subtraction, it should follow the number which has least number of digits after the decimal point.

Now in rounding off numbers we have to consider the number to be dropped so that we can decide if the preceding digit can be retained or should be increased by 1.  If the number to be dropped is 5 and above, the preceding digit should be increased by 1 but if the number to be dropped is less that 5 the preceding digit should be retained.  In the above example, since there should be 1 digit after the decimal point, the number to be dropped is 3 which is less than 5 and so we retain the value of the preceding digit which is 4.

2.  In multiplication and division, the calculated result should be rounded off to proper number of significant figures following the value or number which has least number of significant figures.  Let us have an example:

2.8  x  4.5039  =  12.61092    or 13

The above example was rounded off to 13 having only two significant figures following the least number of significant figures from the given,  two significant figures for 2.8 and 5 significant figures for 4.5039.  

6.85  ÷  112. 04 = 0.061138879 or 0.0611

The calculation above was rounded off to 3 significant figures following the number of significant figures of 6.85.

Try this!

Solve the problem below and round off to proper number of significant figures.

1.  11,254.1 g + 0.1983 g

2.  66.59 L -  3.113 L

3.  8.16 m x  5.1355 

4.  0.0154 kg ÷ 88.3 mL

Solution:

 1.   11,254.1 g

     +         0.1983  g

     __________

       11, 254.2983 g should be rounded off to 11,254.3 g

2.   66.59 L

      - 3.113 L

    _________

      63.477 L and should be rounded off to 63.48 L

3.  8.16 m x 5.1355   =  41.90568 m  or 41.9 m

4.  0.0154 kg÷ 88.3 mL  =  0.00017444054 kg/mL   or  0.000174 kg/mL

You can also change your answer in number 4 to scientific notation 1.74 x 10^-4 kg/mL.

     

Significant Figures

In calculating problems related to science or when measuring something, it is important that everyone knows significant figures.  Significant figure is a number that contains certain digits and one uncertain digit. In measuring, we approximate the last digit that is why it is called uncertain digit, at the same time when calculating we round off the last digit that is why it is uncertain digit.

Guidelines in Identifying Significant Figures

In scientific work we must be always careful in writing the proper number of significant figures.  Below are the rules in identifying significant figures.

1.  Any digit that is not zero is a significant figure.  Let say for example 315 cm shows 3 significant figures, 4,367 has 4 significant figures.

2.  Zeros in between nonzero digits are significant.  Example 2002 m has 4 significant figures, 40, 505 g has 5 significant figures.  Zeros become significant because it is in between nonzero digits.

3.  Zeros to the left of the first nonzero digit are not significant.  They are only used to determine the placement of the decimal point.  For example, 0.007 g has only one significant figure and 0.00034 has two significant figures.

4. If a number is greater that 1, zeros to the right are significant if it contains decimal point.  Example, 1.0 m has two significant figures, 20.000 cm has five significant figures.  And if the number is less that 1 only zeros to the right are significant.  For example, 0.000240 g has 3 significant figures, 0.000300 m has also 3 significant figures.

5.  For a number that does not have decimal point, the trailing zeros to right may or may not be significant.  Thus if we have 300 cm, there can be 1, 2 or 3 significant figures.  Now to avoid ambiguity, we can write the number in scientific notation.  The number 300 can be written as 3 x 10², which has 1 significant figure only, it can also be 3.0 x 10², this time has 2 significant figures and 3.00 x 10², can have 3 significant figures.


Example: 
Determine the number of significant figures in the following measurements:
a.  5.02 g
b.  0.03459 cm
c.  478 kg
d.  85.0 L
e.  1.330 x 10²² atoms

Solution:
a. There are  3 significant figures because zeros in between nonzero digits are significant,
b. There are 4 significant figures, zeros to the left are not significant.
c.  There are 3 significant figures, all non zero digits are significant.
d. Also has 3 significant figures, zeros to the right are significant if there is decimal point.
e.  There are 4 significant figures, zero to the right is significant because the number contains decimal point.


For addition, subtraction, multiplication and division of significant figures and rounding off the answers to proper number of significant figures just click HERE.

Factors that Affect Chemical Equilibrium

Chemical equilibrium means that there is a balance between the rate of forward and backward reaction.  But this condition is so sensitive that whatever changes in some factors or variables there is a change in the production of products, it can either produce more or less products.  These variables or factors are concentration, pressure, volume, and temperature.

Le Chatelier's Principle

Now in chemical equilibrium there is a general rule that will guide us in predicting the direction in which a chemical reaction will move in case concentration, pressure, volume and temperature are changed, this is called the Le Chatelier's Principle.  This was formulated by a French Chemist Henri Le Chatelier.  Le Chateliers Principle states that if an external stress is applied to a system at equilibrium the system adjust in such a way that the stress is partially offset or balance as it tries to reestablish the equilibrium.  Stress means the variables or factors that are changed, the concentration, pressure, volume and temperature.

Concentration

Concentration is the amount of solute that is present in a given solvent.  The greater the amount of the solute in the solution the higher is the concentration.  Now what happens when there is a change in the concentration of the reactants and products? Let us give an example:

FeSCN⁺²(aq)        ⇄        Fe⁺³(aq)    +   SCN^-(aq)

                                                 red                       pale yellow          colorless

The above reaction shows what happens with Iron (III) thiocyanate [Fe(SCN)3] when dissolve in water.  The solution gives a red color due to the presence of hydrated FeSCN⁺² ion.  The substance when dissolve in water gives Fe⁺³ ion and SCN^- ion.  The system of chemical equilibrium occurs.

Now, what will happen if we add substance into the solution?  Let us say sodium thiocyanate (NaSCN) is added to the solution.  In this case, the stress added to the system is the  increase in the concentration of SCN^- ion from the dissociation of NaSCN.  To offset the stress the Fe⁺³ ions will react with SCN^- ions, and the equilibrium will shift from right to left.

FeSCN⁺²(aq)     ←     Fe⁺³(aq)    +   SCN^-(aq)

The above reaction will deepen the red color.  What if iron (III) nitrate [Fe(NO₃)₃] is added?  The red color will also deepen because the concentration of Fe⁺³ ions will increase from the dissociation of Fe(NO₃)₃, and it will react with SCN^- ions, and therefore the equilibrium will shift from right to left.  Both Na⁺ ions and NO₃^- ions are colorless spectator ions.

Suppose we add oxalic acid (H₂C₂O₄) to the original solution.   Oxalic acid ionizes in water to form oxalate ion (C₂O₄⁺²), which binds strongly with Fe⁺³ ions forming a yellow ion Fe(C₂O₄)₃^-3 which removes free Fe⁺³ ions in solution. And therefore, more FeSCN⁺² units dissociate and the equilibrium shifts from left to right.  The solution will now turn to yellow because of the formation of Fe(C₂O₄)₃^-3.

FeSCN⁺²(aq)     →    Fe⁺³(aq)    +   SCN^-(aq)

Changes in Pressure and Volume

Not all solutions are affected of the change in pressure and volume.  Only gases  are the one affected because of the ability of gases to be compressed.  While liquids and gases are virtually incompressible.   Let us recall the formula for ideal gas equation:

PV = nRT

P = (n/V)RT  

The equation tells that P is inversely related to V, which means as the volume is decreased the pressure increases and vice versa.  On the other hand, the concentration, n, of the gas is directly proportional to the pressure of gas, which means if the concentration is increased the pressure also increases.

Let us have an example, suppose we have the reaction below:

N₂O₄(g)    ⇄    2NO₂(g)

The equilibrium system is in a cylinder fitted with a movable piston.  What will happen if the pressure on the gases is increased?  Increasing the pressure will result in decrease in volume and so concentration of N₂O₄ and NO₂ will increase.  Therefore, increasing pressure will cause the shift of equilibrium to the left, more N₂O₄ will form because of the decrease in volume.  Why N₂O₄?  Simply because N₂O₄ has less number moles.  In other words if pressure is decreased the equilibrium will shift to the right in favor of more number of moles NO₂.

Therefore, an increase in pressure (decrease in volume) favors the net reaction which has less number of moles, in the above reaction, it favors the backward reaction; and the decrease in pressure (increase in volume) favors the the net reaction which has more number of moles.  In the above reaction, it favors the forward reaction.

Let us have another examples:
Predict the direction of the net reaction of the following equilibrium system if pressure will be increased.
a.   2PbS(s)  +   3O₂     ⇄      2PbO(s)   +   2SO₂(g)
b.  PCl₅(g)     ⇄      PCl₃(g)   +   Cl₂(g)

Solution:  
We have to be reminded that only gases are affected by pressure because solids and liquids are incompressible.
a. There are 3 moles of reactants and 2 moles of the products not including the solids, therefore increasing pressure or decreasing the volume will cause the system to shift to the right  in favor of less than number of moles or to the product side.
b.  Here only one mole in the reactant side and two moles in the product side increasing the pressure will shift the equilibrium to the left to the reactant side with less number of moles.

Changes in Temperature

A change on concentration, pressure and volume does not alter the equilibrium constant it only alters the relative amounts of reactants and products, only change in temperature can alter the equilibrium constant.  There are two reactions possible the exothermic reaction and the endothermic reaction.  Exothermic reaction is a chemical reaction that releases heat or energy while endothermic reaction is the chemical reaction that absorbs heat.

Let us have an example,

N₂O₄(g)      ⇄     2NO₂(g)

The above reaction is endothermic in the forward reaction (absorbs heat, △H^o > 0),

heat  +  N₂O₄(g)    →   2NO₂(g)          △H^o = 58.0 kJ/mole

so the reverse reaction is exothermic (releases heat, △H^o < 0)

2NO₂(g)   →    N₂O₄(g)  +  heat      △H^o = -58.0 kJ/mole

At equilibrium, the heat effect is zero because there is not net reaction.  Increasing the temperature will favor the endothermic reaction, the forward reaction (the shift is from left to right), the amount of N₂O₄ decreases and an increase in the amount of NO₂. On the other hand a decrease in the temperature will favor the exothermic reaction reaction, the backward reaction (the shift is from right to left), which decreases the NO₂ and increases N₂O₄.

Therefore we can say that an increase in temperature will favor the endothermic reaction and the decrease in temperature favors the exothermic reaction.

The Effect of Catalyst

Catalyst enhances the chemical reaction. Adding catalyst to the reaction will not alter the chemical equilibrium constant or will not shift the equilibrium to any direction, it just increases the reaction rate by lowering the activation energy of the reacting molecules.

Adding a catalyst to a reaction that is not at equilibrium will only help the reaction to achieve equilibrium at the soonest possible time.  The same equilibrium can be obtained without a catalyst only that it takes longer time to occur.

Ways of Expressing Equilibrium Constants

In my previous post, I posted how to express the Equilibrium Constant Expressions for a given reaction, the standard format using molarity as unit for the concentration of reactants and products.  But we can't avoid the fact that we have different types of reactions, there are reactions which involve the same reacting species and involve all gases, and there are some the same reacting species but involve aqueous solutions. There are also reactions that the reacting species are of different phases.  This post will introduce to you the different ways of expressing equilibrium constant expression.

Homogeneous Equilibria

Homogeneous equilibrium applies to the reaction where the reacting species are in the same phase. Let us have an example: 

In the above reaction, it shows the dissociation of N2O4; the reactant is a gas and the product is a also a gas. We can express two equilibrium constant expressions in this type of reaction:

Kc denotes that the reacting species is expressed in molarity or mole per liter.  Since the reactant and the product in the above reaction are all gases the concentration can also be expressed in terms of their partial pressures.  In gases, at constant temperature the pressure P of a gas is directly related to the concentration in moles per liter; that is P=(n/V)RT.  Thus the above equation, the equilibrium constant can also be expressed using the partial pressures of gases.

PNO2 and PN2O4 are the equilibrium partial pressures (in atmospheres) of NO2 and N2O4 respectively.  On the other hand, the subscript in Kp tells us the equilibrium is expressed in terms of pressures.

Kc is not equal to Kp because the concentration of the reactants and the products are not the same with the partial pressures of the reactants and products.  The relationship between Kc and Kp can be derived using the ideal gas equation PV=nRT.  Since : 

and so Kc and Kp are related 

where 

               0.0821 L.atm/K.mol.  is the universal gas constant (R).  To use the above equation, the pressures in Kp must be in atm.  

In general Kc and Kp are not equal but in special case that the change in n is equal to zero therefore Kc = Kp.

And with regards to the unit of Kc and Kp, since the units are canceled out. there is no unit for K.

Let us have an example.  Write the Kc or Kp expressions of the following reactions if applicable:

We need to keep in mind the following:

     a.  Kp expression applies only to gaseous reaction.

     b.  The concentration of the solvent (particularly water) does not appear in the equilibrium constant expression.

Solution:

a.  

b.  

c.   

Heterogeneous Equilibria

Heterogeneous equilibrium is a reversible reaction involving reactants and products that are in different phases.  Example is the decomposition of calcium carbonate when heated, below is the equilibrium reaction:

There are three separate phases in the above reaction, two solid phases and one gas phase.  In writing equilibrium constant, not only pure liquid which is not included in the equilibrium constant but also pure solids.  Therefore the equilibrium constant expression of the above reaction is:

This was taken from

Since CaO and CaCO3 are solids, the concentration can be converted to 1, and so the final equilibrium constant expression is 

Kp can also be used since CO2 is a gas, therefore  Kp = PCO2.

Sample Problem:

Calculate the Kp and Kc of the reaction below at 800 degree Celsius. The pressure of CO2 is 0.236 atm.

                                       

Solution:

                  Kp = PCO2

                        =  0.236

so,

                  

            

In this case T = 800 + 273 = 1073 K, 🔺n =1. Substituting values, 

                

       

   

The Equilibrium Constant

I discussed in my previous post about chemical equilibrium, in which it is considered a dynamic process, wherein the two processes involved do not stop, they  occur at equal rate.

Two chemists in the name of Cato Guldberg and Peter Waage observed that there is a change in the mass of a substance participating in a reversible reaction produced a shift in equilibrium.  This principle is called the law of mass action.  Chemists studied the following type of reversible reaction:

They experimentally observed that the ratio of the product concentration raised to the powers of c and d to the reactant concentration raised to the powers of a and b, always got a constant value.  This is called the equilibrium constant, keq. Equilibrium constant (Keq) expression of the above reaction is written below:

where [A], [B], [C], [D] are the molar concentrations of reactants and products while a, b, c, and d are the coefficients in the reactants and products.  The above equation is the mathematical form of the law of mass action.   The relationship that is applied to every reversible reaction is called law of chemical equilibrium.

Let us have an example of writing the chemical equilibrium expression of some reversible reaction,

In writing the equilibrium constant expression, keq,  the concentration of the product raise to certain exponent is written as the numerator while the concentration of the reactant raise to a certain power is the denominator.  The exponent is taken from the coefficient of the reactants and products.  Therefore the chemical equilibrium constant expression of the equation above is:

To calculate the equilibrium constant, the concentration of the products are multiplied if there are two or more products and it is being divided to the product of the concentration of the reactants.  When the Keq value is greater than 1, it means that the equilibrium lies to the right and if the Keq value is less than 1 it means that the equilibrium lies to the left.

Let us try some more examples, write the equilibrium constant expression of the following:

Answer:

The Concept of Chemical Equilibrium

Chemical reactions are everywhere.  Some go in one direction only, the forward reaction and some go in two directions, the forward and backward reaction.  The latter is called reversible reaction and it is indicated by a double arrow.  Example, when a glass of water with cover is placed under the sun, it undergoes reversible reaction, the vaporization and condensation processes.  When the rate of vaporization process equalizes with the rate of condensation process, the equilibrium is reached.

Chemical equilibrium is a dynamic process.  It occurs when the rate of forward reaction and the backward reaction are equal and the concentration of the reactants and products no longer change with time.  Chemical equilibrium reaction involves different substances as reactants and products. Equilibrium involving two phases of the same substance just like the example given above, the vaporization and condensation processes is an example of physical equilibrium, because the changes that occurs are only physical processes.

Example of chemical equilibrium is the reversible reaction of nitrogen dioxide (NO2) and dinitrogen tetroxide (N2O4) 

The progress of reaction is easily monitored because there is a corresponding change in the color of the gases, N2O4 is a colorless gas while NO2 has dark brown color.  Let us say for example, an evacuated glass is injected with known amount of N2O4, and suddenly the color changes to dark brown, it means there is already a formation of NO2 gas.  The color intensifies until such time that equilibrium is reached.  Equilibrium is reached when there is no changed in color observed inside the evacuated glass

Advanced pH and pOH Calculations

In my previous post you learned how to calculate the pH and the [H+] of solution even without the calculator using the formula

But what about if the given hydrogen ion concentration is not base of 1 like for example 0.00024 M and the pH is not a whole number like for example 7.5 pH.  You might say, use calculator.  Yes you are right calculator can be used.  But this post will calculate pH and [H+] without the use of calculator.

Converting [H+] to pH

In this type of calculation the same formula will be used only that additional steps will be added.

Example,  a solution has a hydrogen concentration, [H+]  of  0.00015 M, what is the pH of the solution?

Solution:

The pH is expression is 

We have to recall some rules in logarithm.  The logarithm of a product is equal to the sum of the logarithm of each value.  That is, log a x b = log a + log b .  Applying rule to the above equation

Now its time to use logarithmic table as reference table for the logarithm of different numbers or you can have list of the logarithmic value of different numbers researched from the internet.  

Continuing the calculation,

This means that if a solution has hydrogen ion concentration of 0.00015 M has a pH of 3.82.

Another example: 

    

Stomach acid has a hydrogen ion concentration, [H+] of  0.020 M.  What is the pH of the stomach acid?

Solution:   

Converting pH to [H+]

In my previous blog [H+] is calculated using 

Let us see for example the pH is 3.80,

We have to remember the rule of exponent, to  multiply two exponential number, add the powers together. That is, .  Let us apply the rule on the above pH.

3.80 is the same as 0.20 + (-4).  

This time we need to use the antilogarithm table, 0.20 has antilog of 1.6, therefore 

This means that pH equal to 3.80 has a concentration of 0.000 16 M.

Another example:

Calculate the [H+] of seawater which has a pH of 7.85.

Solution:

pOH

What is pOH?  pOH expresses the basicity of the solution.  It is called the power of hydroxide.   The formula is the same with that of pH.  

The calculation is the same with pH and the [OH-].  Below is the summary between pH, pOH.

TRY THIS:

Calculate the pH of the following  [H+]

1.  0.000 000 047 M

2.  0.000 089 M

3.  0.000 003 7 M

Calculate the [H+] given the following pH

1.  pH = 3.25

2.  pH = 8.68 

3.  pH = 6.20

The pH Concept

In my previous post, I posted the definition of acids and bases and the properties of acids and bases. We learned that acids donates H+ ions in solution while bases donates OH- ions in solution.  pH is defined as the measure of acidity and basicity of substance in solution.  pH also measures the  H+ ion concentration in solution.  The range goes from 0 - 14 where pH 0-6 indicates that a solution is acidic,  pH 7 means neutral substances and pH 8-14 indicates basic substances.

Converting [H+] to pH

The pH scale expresses the molar hydrogen ion concentration [H+], as a power of 10.  It means pH is the negative logarithm of the molar hydrogen ion concentration. That is 

Example 1.

What is the pH of the solution if the hydrogen ion concentration, [H+] is 0.1 M?

Solution:

 The easiest way to calculate the concentration in pH is to express the concentration first to the power of 10.  See the calculation below:

The log of  is -1, since we get the - log the pH equals 1.

Example 2.

Calculate the pH of vinegar with hydrogen ion concentration of [H+] 0.001 M.

Solution:

Converting pH to [H+]

We can calculate the hydrogen ion concentration [H+] given the pH by rearranging the equation :

For example, if a solution has a pH of 5, what is the [H+]?  Look at  the solution below:

Just make the value of pH as the exponent and express it in concentration by molarity.

Another example:

Calculate the hydrogen ion concentration of a solution having pH of 4.

Solution:

TRY THIS:

Ionization of Water

We learned in my previous post the different weak electrolytes and strong electrolytes.  Strong electrolytes are 100% dissociated or ionized in water while weak electrolytes only few % are ionized in water.  Example is the salt in water solution, using electrical conductivity apparatus we can tell that its a strong electrolyte because it can produce very bright light while acetic acid on the other hand is a weak electrolyte, it only produce dim light in electrical conductivity apparatus.  Water on the other hand is an example of  weak electrolyte, lets see what happens in when water is combined with another water molecule:

Water molecule reacts with another water molecule forming hydronium ion and hydroxide ion.  The double arrow indicates that a reversible reaction occur, hydronium ion and hydroxide ion can also combine forming water molecules.  This is called autoprotolysis (or autoionization)  of water.  The bond breaking between water molecule and ions combining forming water molecules continues until the same rate happens, and this is called state of chemical equilibrium.

We can write simply the ionization of water molecules as follows:

The concentration of hydrogen ion in pure water is  mole per liter at . 

Since the ionization of water gives one H+ and OH-  , therefore the ionization of hyroxide ion is also  mole per liter at the same temperature.  The product of the concentration of the H+ and the OH- is called the equilibrium constant of water (symbol Kw).  Chemist use bracket [ ] to represents the concentration, therefore [H+] is the concentration of hydrogen ion and [OH-] is the concentration of hydroxide ion.  Kw of water can be calculated using the formula:

Calculating the Kw of water

Based from the calculation above Kw (equilibrium constant) of water is . Since Kw is constant whenever there is a change in the concentration of hydrogen ion the concentration of hydroxide ion is also changes or vise versa.

Example:

Calculate the hydroxide ion concentration [OH-] given the hydrogen ion concentration [H+]:

Solution:

  TRY THIS: